(i) The expression for the displacement of P from O, in terms of t, can be found by integrating the velocity function v(t):

S(t) = ∫(4t² - 2t - 2) dt

Integrating each term of the velocity function:
S(t) = (4/3)t³ - t² - 2t + C

Given that the displacement at t = 0 is 9 meters, we can substitute this information into the equation:

9 = (4/3)(0)³ - (0)² - 2(0) + C
C = 9

So, the expression for the displacement of P from O, in terms of t, is:
S(t) = (4/3)t³ - t² - 2t + 9

(ii) To find when P is at instantaneous rest, we need to find the value(s) of t for which the velocity v(t) is equal to 0:

4t² - 2t - 2 = 0

This quadratic equation can be solved using factoring, quadratic formula, or completing the square. Using the quadratic formula:

t = (-(-2) ± √((-2)² - 4(4)(-2))) / (2(4))
t = (2 ± √(4 + 32)) / 8
t = (2 ± √36) / 8
t = (2 ± 6) / 8

We get two solutions: t = 1/2 and t = -1. We discard the negative value as it is not meaningful in this context.

So, the value of t when P is at instantaneous rest is t = 1/2 second.

(iii) To find the distance of P from O at t = 3, we substitute t = 3 into the displacement function:

S(3) = (4/3)(3)³ - (3)² - 2(3) + 9

Calculating the expression, we get:
S(3) = 36 - 9 - 6 + 9 = 30 meters

Therefore, the distance of P from O at t = 3 is 30 meters.

(iv) The distance traveled by P in the first 6 seconds can be found by calculating the total displacement over that time period. We can find the displacement at t = 6 and subtract the displacement at t = 0:

Distance traveled = S(6) - S(0)

Substituting t = 6 and t = 0 into the displacement function:
Distance traveled = [(4/3)(6)³ - (6)² - 2(6) + 9] - [(4/3)(0)³ - (0)² - 2(0) + 9]
Distance traveled = (4/3)(216) - 36 - 12 + 9 - 9
Distance traveled = 288 - 36 - 12
Distance traveled = 240 meters

Therefore, the distance traveled by P in the first 6 seconds of its motion is 240 meters.